TITLE: Why do some say that $f(x)=\frac{1}{x}$ discontinuous?
QUESTION [5 upvotes]: A continuous function $f:X\to Y$ is one which satisfies the following property: for every open set $U\subset Y$, the set $f^{-1}(U)$ is open in $X$. 
I don't see why according to this definition $f(x)=\frac{1}{x}$ is discontinuous at $0$. The inverse image of every open set of the form $(a,\infty)$ is open, it is $(0,\frac{1}{a})$. Similarly, the inverse image of $(-\infty,-a)$ is $(-\frac{1}{a},0)$.
And one can see that the inverses of other forms of open sets are also clearly open. 
Why then is $1/x$ discontinuous according to some?

REPLY [6 votes]: For what's worth, I checked two french books and Rudin's PMA, and in all cases, discontinuity is only defined for a point of the domain of definition of the function. As the other answers suggest, there may be other definitions of "discontinuity", for points where $f$ is not defined. I would say it's relatively harmless, apart from introducing misleading terminology.

It's misleading for another reason than just terminology, I think. A function is defined to be given by a set of "inputs", a set of "outputs", and a relation such that each input has an output. It makes little sense to tell anything about (continuity of) a function where it's not defined, in some sense it doesn't even "belong" to the function.
For example, $f:x\rightarrow 1/x$ is defined on $]-\infty,0[\;\cup\;]0,+\infty[$. I could tell it's discontinuous at $0$. But then I could also tell it's discontinuous at $(1,2)$ (where this is a point in $\Bbb R^2$), or discontinuous at $\emptyset$, or anything completely meaningless: any mathematical object that is not part of the set of inputs may be said to be a point of discontinuity, but you learn nothing.
Notice that you can also define a function $g:]1,2[\rightarrow \Bbb R$ such that $g(x)=1/x$. Then $f$ and $g$ are not the same function, even though they are given by the same formula. And even $h:]1,2[\rightarrow \Bbb ]1/2,1[$ such that $h(x)=1/x$ is not the same thing as $g$. For example, $h$ is a bijection, whereas $g$ is not. Thus, the three components (input set, output set, relation) are all very important in the definition of a function.
Here, $0$ has a particular status only because when we see $]-\infty,0[\;\cup\;]0,+\infty[$, we see also the whole real line, and this additional number $0$. But the function is just not defined at $0$. This conveys more useful information to say that, than to say it's discontinuous.

Whatever we may decide about discontinuity where the function is not defined, there is a good news: there won't be another definition of continuity, for one good reason. The topological definition of continuity states that, given two topological spaces $A,B$, $f:A\rightarrow B$ is continuous iff $f^{-1}(S)$ is open for all open subsets $S \subset B$.
This is the standard definition of continuity, and is equivalent in the real case with the definition which implies that $x\rightarrow 1/x$ is continuous. For example, for $a>0$, 
$$f^{-1}(]-\infty,-a[ \;\cup\; ]a,+\infty[)=]-1/a,0[\;\cup\;]0,1/a[$$
which is an open set.

Hence, I would say that these books I checked do the right thing: where the function is defined, we can argue about continuity or discontinuity. Where it's not defined, it's meaningless.